# Differential EQNS: $y \, dx = \left[ x + (y^2 - x^2)^{1/2} \right] dy$

Submitted by Sydney Sales on July 14, 2016 - 10:53am

ydx = x + √ y^2 - x^2 dy

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Submitted by Sydney Sales on July 14, 2016 - 10:53am

ydx = x + √ y^2 - x^2 dy

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- 669 reads

## Please use proper groupings

Please use proper groupings by using parenthesis. Hindi kasi malinaw ang problem.

## ydx= x + ( y^2 - x^2) ^ 1/2

ydx= x + ( y^2 - x^2) ^ 1/2 dy

## The x don't have differential

The x don't have differential element, do you mean this?

$y \, dx = x + (y^2 - x^2)^{1/2} \, dy$

Or this?

$y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$

## ydx = [x + ( y^2 - x^2 )^ 1/2

ydx = [x + ( y^2 - x^2 )^ 1/2] dy

this one.

## Solution

Solution$y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$

dx = v dy + y dv

$y(v\,dy + y\,dv) = \left[vy + (y^2 - v^2y^2)^{1/2} \right] dy$

$vy\,dy + y^2\,dv = \left[vy + y(1 - v^2)^{1/2} \right] dy$

$y^2\,dv = y(1 - v^2)^{1/2}\,dy$

$y\,dv = (1 - v^2)^{1/2}\,dy$

$\dfrac{dv}{(1 - v^2)^{1/2}} = \dfrac{dy}{y}$

$\displaystyle \int \dfrac{dv}{\sqrt{1 - v^2}} = \int \dfrac{dy}{y}$

$\arcsin v = \ln y + c$

$\arcsin \left( \dfrac{x}{y} \right) = \ln y + c$

## sa factoring po. bakit po

sa factoring po. bakit po naging.

[ vy + y (1- v^2 )^1/2 ] dy

## (y2 - v2y2)1/2

(y

^{2}- v^{2}y^{2})^{1/2}= [ y

^{2}(1 - v^{2}) ]^{1/2}= y

^{2(1/2)}(1 - v^{2})^{1/2}= y (1 - v

^{2})^{1/2}## salamat po.

salamat po.

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